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7.35: Untitled Page 180

  • Page ID
    18313
  • Chapter 7

    and choose a value for the parameter, q , that provides for a severely damped convergence. The results are shown in Table 7.8b where we see that the iterative procedure converges rapidly to the value given by x  ( M

    )

      0 . 6108

    (42)

    C2H4 5

    Use of this result in Eq. 20 allows us to determine the fraction of Stream #4

    that is purged and this fraction is given by

    ( M

    ) CY

    C H

    5 

      1 

    (43)

    2 4 C

     0 2874

    .

    1

    which is identical to that obtained in Example 7.6.

    Table 7.8b. Iterative Values for Dimensionless Flow Rate

    (Wegstein’s Method)

    i

    q

    xi

    xi+1

    0

    0.9950

    0.7500

    0.4070

    1

    0.9950

    0.4070

    0.5265

    2

    0.9950

    0.5265

    0.5938

    3

    0.9950

    0.5938

    0.6109

    4

    0.9950

    0.6109

    0.6108

    5

    0.9950

    0.6108

    0.6108

    7.5 Problems

    Note: Problems marked with the symbol  will be difficult to solve without the use of computer software.

    Section 7.1

    7‐1. In the production of formaldehyde ( CH O ) by catalytic oxidation of 2

    methanol ( CH OH ) an equi‐molar mixture of methanol and air (21% oxygen 3

    and 79% nitrogen) is sent to a catalytic reactor. The reaction is catalyzed by finely divided silver supported on alumina as suggested in Figure 7.1 where we

    index-346_1.png

    Material Balances for Complex Systems

    337

    Figure 7.1. Production of formaldehyde

    have indicated that carbon dioxide ( CO ) is produced as an undesirable 2

    product. The conversion for methanol ( CH OH ) is given by 3

     RCH OH

    C  Conversion of CH OH 

     0 2

    . 0

    3

    3

    M CH3OH1

    and the selectivity for formaldehyde / carbon dioxide is given by RCH O

    2

    S  Selectivity of CH O/CO

     8 . 5

    2

    2

    RCO2

    In this problem you are asked to determine the mole fraction of all components in the Stream #2 leaving the reactor.

    7‐2. Use the pivot theorem (Sec. 6.4) with Eq. 6 of Example 7.1 to develop a solution for R

    and R

    using ethylene ( C H ) , methane ( CH ) and

    H2

    C2H6

    2

    4

    4

    propylene ( C H ) as the pivot species. Compare your solution with Eq. 7 and 3

    6

    Eq. 8 of Example 7.1. In order to use ethylene ( C H ) , ethane ( C H ) , and 2

    4

    2

    6

    propylene ( C H ) as the pivot species, one needs to use a column/row 3

    6

    interchange (see Sec. 6.2.5) with Eq. 6 of Example 7.1. Carry out the appropriate column /row interchange and the necessary elementary row operations and use the pivot theorem to develop a solution for R

    and

    H

    R

    .

    CH

    2

    4

    7‐3. An industrial process for making acetic anhydride utilizes the highly unsaturated and reactive ketene ( CH CO ) that is also an important intermediate 2

    in other chemical processes. The pyrolysis of acetone ( CH COCH ) in an 3

    3

    externally heated empty tube produces ketene ( CH CO ) and methane ( CH ) .

    2

    4

    This is illustrated in Figure 7.3 where we have indicated that some of the ketene

    index-347_1.png

    index-347_2.png

    index-347_3.png

    338